Subnetting Questions 33
Enter the first valid host on the network 172.19.96.0 255.255.224.0:
Solution
Hence, the IP Address starts from 172 it indicates it is a Class B Address. By default, in Class B address 16 bits are reserved for the network and 16 bits are reserved for the host.
Hence the Subnet Mask is 255.255.224.0. We have to borrow 3 bits from the third octet (27 + 26 + 25 = 224).
(27 + 26 + 25 + 24 + 23 + 22 + 21 + 20) . (27 + 26 + 25 + 24 + 23 + 22 + 21 + 20) . (27 + 26 + 25 + 24 + 23 + 22 + 21 + 20) . (27 + 26 + 25)
11111111 . 11111111 . 11100000 . 00000000
Let’s count the number of subnets
2n (Here, n indicates the total number of network bits borrowed)
23 = 8 Subnets are available.
Let’s count the number of IP Address on each network
2n (Here, n indicates the total number of host bits)
213 = 8192 IP Address is available on each network
List of Networks with the number of host and broadcast address
| Network Address | Usable Host Range | Broadcast Address |
| 172.19.0.0 | 172.19.0.1 – 172.19.31.254 | 172.19.31.255 |
| 172.19.32.0 | 172.19.32.1 – 172.19.63.254 | 172.19.63.255 |
| 172.19.64.0 | 172.19.64.1 – 172.19.95.254 | 172.19.95.255 |
| 172.19.96.0 | 172.19.96.1 – 172.19.127.254 | 172.19.127.255 |
| 172.19.128.0 | 172.19.128.1 – 172.19.159.254 | 172.19.159.255 |
| 172.19.160.0 | 172.19.160.1 – 172.19.191.254 | 172.19.191.255 |
| 172.19.192.0 | 172.19.192.1 – 172.19.223.254 | 172.19.223.255 |
| 172.19.224.0 | 172.19.224.1 – 172.19.255.254 | 172.19.255.255 |
The network ID 172.19.96.0 is falls in fourth place and its first host is 172.19.96.1